Pie chart not displaying when plotting big and small numbers at the same time.

[baptistelabat-syroco]

When I plot a piechart, if there is a number much larger than another one (by a factor 100 000 000), no piechart is displayed.
If I then unselect one of the value, the piechart is displayed.
If i reduce the factor between the two numbers, the piechart is displayed as well.

Here is a example:

https://jsfiddle.net/3qr50za6/

var data = [
{
"type": "pie",
"values": [
"1",
"1e-8",
],
"labels": [
"Number",
"Much smaller number",
],
}
];

var layout = {
"title": "Small and big numbers issues",
"showlegend": true,
width: 1000,
height: 400
};

Plotly.newPlot('myDiv', data, layout);

[alexcjohnson]

Thanks @baptistelabat-syroco!

Looks like the SVG engine treats the start and end points as identical in this case, so it can't figure out how to draw the nearly-360-degree arc. We already catch this when the arc is exactly 360 degrees:

plotly.js/src/traces/pie/plot.js

Lines 101 to 111 in f316a7a

	if(pt.v === cd0.vTotal) { // 100% fails bcs arc start and end are identical
	var outerCircle = 'M' + (cx + pt.px0[0]) + ',' + (cy + pt.px0[1]) +
	arc(pt.px0, pt.pxmid, true, 1) +
	arc(pt.pxmid, pt.px0, true, 1) + 'Z';
	if(hole) {
	slicePath.attr('d',
	'M' + (cx + hole * pt.px0[0]) + ',' + (cy + hole * pt.px0[1]) +
	arc(pt.px0, pt.pxmid, false, hole) +
	arc(pt.pxmid, pt.px0, false, hole) +
	'Z' + outerCircle);
	} else slicePath.attr('d', outerCircle);

Seems like we just need to loosen this criterion. Might as well do this for any slice bigger than 50%. Should be as simple as changing pt.v === cd0.vTotal to pt.v > cd0.vTotal / 2 and changing a couple of those pt.px0 to pt.px1 (probably the last one in the main arc and the first in the hole)... then adding the condition you found to a test image. Feel like making a PR???

[baptistelabat-syroco]

Hi alexcjohnson, I am coding more in python than javascript, so not sure how to handle and test this.

We have a workaround in python, but better to correct it at root.
values=np.round(np.array(values) / max(values), 6)*max(values)

I'll let you know if I can find more time to invest or find support for this.